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Division - Decimals.

133. Find the quotient of 785.6 divided by .5. Operation.

Explanation. .5)785.6'5 First place a separatrix (~) after that figure in 1571.3

the dividend that is of the same denomination as

the right hand figure of the divisor — in this case, after the figure 6. Then divide, writing the decimal point in the quotient when, in the process of division, the separatrix is reached in this case, after the figure 1.

It was required to find how many times 5 tenths are contained in 7856 tenths. 5 tenths are contained in 7856 tenths 1571 times. There are yet 15 hundredths to be divided. 5 tenths are contained in 15 tenths, 3 times; in 15 hundredths, 3 tenths of a time.

NOTE. — By holding the thought for a moment upon that part of the dividend which corresponds in denomination to the divisor, the place of the decimal point becomes apparent at once.

5 apples are contained in 7856 apples, 1571 times.
5 tenths are contained in 7856 tenths, 1571 times.

134. Solve and explain the following problems with special reference to the placing of the decimal point :

1. Divide 340 by .8.
2. Divide 468.5 by .25.
3. Divide 38.250 by 12.5.
4. Divide 87 by 2.5.
5. Divide 546 by .75.
6. Divide .576 by 2.4.

.8)340.00 .25)468.50V 12.5)38.2°50

2.5)87.01 .75)546.00

2.4).5'76 8. 94.5+.8 10. 125.5

7. 86+.375
9. 75:.15

(a) Find the sum of the ten quotients.

(P. 327.)

PROBLEM.

Division - United States Money. 135. Divide $754.65 by $.27. Operation.

Explanation. $.27)$ 754.652795 This means, find how many times 27 54

cents are contained in 75465 cents. 27 214

cents are contained in 75465 cents, 2795 189

times. 256

At 274 a bushel, how many bushels of 243

oats can be bought for $754.65 ? As many 135

bushels can be bought, as $.27 is contained 135

times in $ 754.65. It is contained 2795 times. Therefore, 2795 bushels can be

bought. 136. Divide $ 754.65 by 27. Operation.

Explanation. 27)$754.'65($27.95 This means, find one 27th of $754.65. 54

One 27th of $754.65 is $27.95. 214

NOTE. — One might find one 27th of 189

$754.65 by finding how many times $27

is contained in $754.65. See p. 52, Note. 256 243 135

If 27 acres of land are worth $754.65, 135

how much is one acre worth?

PROBLEM.

137. Divide $754.65 by .27. Operation.

Explanation. .27)$ 754.65%$ 2795 This means, find 100 27ths por $754.65. 54

One 27th of $754.65 is $27.95. 10Q 27ths 214

of $754.65 is $2795. 189

NOTE. — In practice we find one 27th

of 100 times $754.65.
256
243

PROBLEM.
135

If.27 of an acre of land is worth $754.65, 135

how much is 1 acre worth at the same rate ?

(P. 328.)

DENOMINATE NUMBERS.

138. Divide 46 rd. 12 ft. 8 in. by 4. Operation.

Explanation. 4)46 rd. 12 ft. 8 in. This means, find i fourth of 46 rd. 12 ft. 11 rd. 11 ft. 5 in.

8 in.

One fourth of 46 rd. is 11 rd. with a remainder* of 2 rd. ; 2 rd. equal 33 ft. 33 ft. plus 12 ft. equal 45 ft.

One fourth of 45 ft. equals 11 ft. with a remainder of 1 ft. ; 1 ft. equals 12 in.; 12 in. plus 8 in. equals 20 in. One fourth of 20 in. equals 5 in.

One fourth of 46 rd. 12 ft. 8 in. equals 11 rd. 11 ft. 5 in.

PROBLEM.

The perimeter of a square garden is 46 rd. 12 ft. 8 in. How far across one side of it ?

139. MISCELLANEOUS. Tell the meaning of each of the following, solve, explain, and state in the form of a problem the conditions that would give rise to each number process.

1. Multiply 64 rd. 14 ft. 6 in. by 8. 2. Divide 37 rd. 15 ft. 4 in. by 5. 3. Divide $675.36 by $48. 4. Divide $675.36 by 48. 5. Divide $675.36 by .48. 6. Divide $675.36 by $4.8. 7. Divide $675.36 by 4.8. 8. Divide $675.36 by $.48. 9. Multiply $356.54 by .36. 10. Multiply $356.54 by 3.6. 11. Multiply $356.54 by 36. 12. Multiply $275.56 by 2.25. 13. Multiply $275.56 by 27. 14. Can you multiply by a number of dollars ? 15. Can you divide by a number of dollars ? (P. 329.)

* The word remainder in this connection suggests incomplete division. When the division is complete there can be no remainder.

[blocks in formation]

1. Observe that in the above examples we divide each term of the dividend by the divisor.

2. Prove Nos. 1 and 3, by (1) reducing each dividend to its simplest form, (2) dividing it, so reduced, by the divisor, and (3) comparing the result with the quotient reduced to its simplest form.

3. Verify No. 2, by letting a= 3, and b= 5.
4. Verify No. 4, by letting a=3, b=5, and c= 7.

141. (6 xa xa xa xa xa) · (2 xa xa)=6a5 = 2 aạ=3 a3.

Observe that to divide one algebraic term by another we must find the quotient of the coefficients and the difference of the exponents.

142. PROBLEMS. 1. 6 a5b 2 a=

3. 8 a3 1,3 = 2 a= 2. 4 a462 : 2 a=

4. 10 a 14 : 2 a=

5.
2 a)6 a5b + 4 a4b2 – 8 a3b3 + 10 a284

6. Verify Problem 5, by letting a = 3, and b= 5.

Algebraic Division.

143. PROBLEMS. 1. Divide 4 aor + 8 a-ra + 6 ar: by 2 ax. 2. Multiply the quotient of problem 1, by 2 ax. 3. Verify problems 1 and 2, by letting a= 2, and x=

r=3.

4. Divide 3 abs + 6 a62 +9 aob by 3 ab. 5. Multiply the quotient of problem 4, by 3 ab. 6. Verify problems 4 and 5, by letting a=3, and b=5.

7. Divide 2 x3y + x2y2 – xy3 by xy.
8. Multiply the quotient of problem 7 by xy.
9. Verify problems 7 and 8, by letting x = 2, and y= 3.

10. Divide 5 aoy2 — 2 a ́y3 + a2y4 by a-y. 11. Multiply the quotient of problem 10, by aạy. 12. Verify problems 10 and 11, by letting a=1, and y=2.

13. Divide 3 b4x + 63x2 – 3 62x3 by bx.
14. Multiply the quotient of problem 13, by bx.

15. Verify problems 13 and 14, by letting b=3, and x= 4.

Observe that when the divisor is a positive number, each term of the quotient has the same sign as the term in the dividend from which it was derived. 2)8 – 6 One half of + 8 is + 4; one half of – 6 is 3.

4-3

16.
2 x)4 x6 – 6 x4 + 8 +3 – 2 x2 + 6 x.

17. Verify by letting x= 2.

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